Characteristic Polynomial of a Matrix

Let $A$ be a $n \times n$ matrix over some field $k$ The characteristic polynomial $p_A(x)$ of $A$ in an indeterminate $x$ is defined by the determinant:

$$ p_A(x):=\det(A-x I) = \left|\begin{matrix} a_{11}-x & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22}-x & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}-x \end{matrix} \right|$$

Remarks

• The polynomial $p_A(x)$ is an $n$ -degree polynomial over $k$
• If $A$ and $B$ are similar matrices, then $p_A(x)=p_B(x)$ because \begin{eqnarray*} p_A(x) &=& \det(A-xI) = \det(P^{-1}BP-xI) \\ &=& \det(P^{-1}BP-P^{-1}xIP) = \det(P^{-1})\det(B-xI)\det(P) \\ &=& \det(P)^{-1}\det(B-xI)\det(P)=\det(B-xI) = p_B(x) \end{eqnarray*}for some invertible matrix $P$
• The characteristic equation of $A$ is the equation $p_A(x)=0$ and the solutions to which are the eigenvalues of $A$

Characteristic Polynomial of a Linear Operator

Now, let $T$ be a linear operator on a vector space $V$ of dimension $n<\infty$ Let $\alpha$ and $\beta$ be any two ordered bases for $V$ Then we may form the matrices $[T]_{\alpha}$ and $[T]_{\beta}$ The two matrix representations of $T$ are similar matrices, related by a change of bases matrix. Therefore, by the second remark above, we define the characteristic polynomial of $T$ denoted by $p_T(x)$ in the indeterminate $x$ by $$p_T(x):=p_{[T]_{\alpha}}(x).$$ The characteristic equation of $T$ is defined accordingly.